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Sonja [21]
3 years ago
13

What is the friction​

Physics
1 answer:
Mnenie [13.5K]3 years ago
3 0

Explanation:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in

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A bowling ball rolls off the edge of a cliff, moving horizontally at 20 m/s. I have to plot the position of the bowling ball on
liubo4ka [24]
For counting x you use simple equation for the distance covered by the object when it moves with constant velocity:
s=v*t
that gives you 20m after 1st second, 40 m after 2nd second, 60 m after 3rd second and so on.

For counting y you have to use the equation for the distanced covered by the object moving with constantly accelerating velocity (symbols refering to vertical movement):
h=g\frac{t^{2}}{2}
that gives you 5m after 1st second, 20m afters 2nd second, 45m after 3rd second and so on.
Add minus signs before y positions to receive graph presenting the movement of the ball.
So the points are: P1=[20,-5], P2=[40,-20], P3=[60,-45] and so on... Pn=[x,y].
8 0
3 years ago
Can someone help check question 2 I don’t think part 2 is right?!
Alisiya [41]
Your answer for question two is right
4 0
3 years ago
he magnetic flux through a loop of wire decreases from 1.7 Wb to 0.3 Wb in a time of 0.4 s. What was the average value of the in
Pani-rosa [81]

Answer:

Induced emf through a loop of wire is 3.5 V.

Explanation:

It is given that,

Initial magnetic flux, \phi_i=1.7\ Wb

Final magnetic flux, \phi_f=0.3\ Wb

The magnetic flux through a loop of wire decreases in a time of 0.4 s, t = 0.4 s

We need to find the average value of the induced emf. It is equivalent to the rate of change of magnetic flux i.e.

\epsilon=-\dfrac{\phi_f-\phi_i}{t}

\epsilon=-\dfrac{0.3\ Wb-1.7\ Wb}{0.4\ s}

\epsilon=3.5\ V

So, the value of the induced emf through a loop of wire is 3.5 V.

5 0
4 years ago
8) A plastic rod, initially uncharged, is rubbed with wool and obtains a charge of 10 C. What is the charge on the wool after ru
LenKa [72]

Answer:

The charge on the wool after rubbing is - 10 C

Explanation:

Every uncharged body is electrically neutral, if the plastic rod acquires 10 Coulombs of charge after been rubbed with wool, then the wool will be left with an equal but opposite charge. This shows that the initial charge on the wool is 10 protons and 10 electrons and when the plastic acquires 10 C (10 protons), the wool will be left with excess 10 electrons.

Therefore, the charge on the wool after rubbing is - 10 C (negative 10 Coulombs).

6 0
4 years ago
A 8.0-cm-diameter horizontal pipe gradually narrows to 5.0 cm . When water flows through this pipe at a certain rate, the gauge
adelina 88 [10]

Answer:

A 8.0 cm diameter horizontal pipe gradually narrows to 5.0 cm. The the water flows through this pipe at certain rate, the gauge pressure in these two sections is 31.0 kPa and 24.0 kPa, respectively. What is the volume of rate of flow?

The flow rate is 3.1175×10⁻³ m³/s

Explanation:

To solve the question we rely on Bernoulli's principle as follows P_{1} +\frac{1}{2}\rho v^{2} _{1} + \rho gz_{1} = P_{2} +\frac{1}{2}\rho v^{2} _{2} + \rho gz_{2}

thus where the pipe is  horizontal we have

z₁ = z₂ hence the above equattion becomes

P_{1} +\frac{1}{2}\rho v^{2} _{1}  = P_{2} +\frac{1}{2}\rho v^{2} _{2}

since the flow rate is constant then

Q = v₁A₁ = v₂A₂

Where is the area of the two sections given by A₁ = π·D₁²÷4 and

A₂ = π·D₂²÷4

Thereffore A₁ = π·0.08²÷4 = 5.02×10⁻³ m²

and A₂ = π·0.05²÷4 = 1.96×10⁻³ m²

v₁ = v₂A₂/A₁ =0.391×v₂

The given pressures are P₁ = 31.0 kPa and P₂ = 24.0 pKa and

ρ = 1000 kg/m³

Plugging the values into the above equation we get

31.0 kPa +0.5× 1000 kg/m³× (0.391×v₂)² = 24.0 pKa +0.5×1000 kg/m³×v₂²

= 31000+76.3·v₂² =24000+500·v₂²

or 423.706·v₂² = 7000

v₂² = 7000/423.706 = 16.52 or  v₂ = 4.065 m/s and  v₁ 0.391×4.065 = 1.59 m/s

The flow rate = v₂A₂ = 1.59×1.96×10⁻³ = 3.1175×10⁻³ m³/s

5 0
4 years ago
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